Question

A 1.50 µF capacitor and a 3.50 µF capacitor are connected in series across a 2.50 V battery. How much charge (in µC) is stored on each capacitor?

Answer 1

Explanation:

The given data is as follows.

      $C_{1} = 1.50 \times 10^{-6} F$

      $C_{1} = 3.50 \times 10^{-6} F$    

      Voltage = 2.50 V

Hence, calculate the equivalence capacitor as follows.

    $\frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$

    $\frac{1}{C} = \frac{1}{1.50 \times 10^{-6} F} + \frac{1}{3.50 \times 10^{-6} F}$

                 = $0.945 \times 10^{-6} F$

          C = $1.06 \times 10^{-6} F$

Now, we will calculate the charge across each capacitance as follows.

              Q = CV

                  = $1.06 \times 10^{-6} F \times 2.50 V$

                  = $2.65 \times 10^{-6} C$

                  = $2.65 \mu C$

Thus, we can conclude that $2.65 \mu C$ is the charge stored on each given capacitor.